blocks|key|1874534|text|在第一种情况下，您将获取指向成员的指针B::b的地址。因为这样的指针不是A父对象的成员，而是一个单独的对象，它不能通过protected机制访问它。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1874535|在第二种情况下，您将询问特定b实例的地址，并使用其基类对其进行限定，以便在多重继承的情况下，编译器将知道您所指的是哪个基类。在此上下文中，protected属性是可见的。|1874536|请注意，这段代码将进行编译：|1874537|class+B
{
protected:
int+b;
};

class+A+:+public+B
{
public:
void+foo(){+&A::b;+}++//+Note+here+&A::+instead+of+&B::
};|code-block|syntax|javascript|1874538|作为一个附加示例，它不能工作的原因与以下(希望更熟悉的)代码不能工作的原因相同：|1874539|class+B
{
protected:
int+b;
};

class+A+:+public+B
{
public:
void+foo(const+B*+b_obj)+{+b_obj->b;+}
};|1874540|entityMap^0|J|4|10|1|0|E|1|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|12|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|13|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|14|8|@]|D|@]|E|$]]|$1|Q|3|R|5|L|7|15|8|@]|D|@]|E|$M|N]]|$1|S|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|T|$]]

In the first case you're taking the address of pointer-to-member <code>B::b</code>. Since such a pointer is NOT a member of the parent of <code>A</code> but a separate object, it can't access it via the protected mechanism.

In the SECOND case where it works you're asking for the address of the specific instance of <code>b</code>, qualifying it with its base class so that in the case of multiple inheritance the compiler would know which base class you mean. In this context the protected attribute is visible.

Note that this compiles:

<pre><code>class B
{
protected:
int b;
};

class A : public B
{
public:
void foo(){ &amp;A::b; } // Note here &amp;A:: instead of &amp;B::
};
</code></pre>

As an added example it doesn't work for the same reason that the following (hopefully more familiar) code doesn't work:

<pre><code>class B
{
protected:
int b;
};

class A : public B
{
public:
void foo(const B* b_obj) { b_obj-&gt;b; }
};
</code></pre>

blocks|key|1042680|text|当您尝试返回值时，这两个语句之间的差异会变得更加明显：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1042681|int*+++++foo()++++{+return+&(B::b);}++//+This+is+a+pointer+to+an+int


int+A::*+foo()++++{+return+&B::b;+}+++//+This+is+a+pointer+to+a+member+of+type+int|code-block|syntax|javascript|1042682|您要做的是通过A对象访问它：|1042683|int+A::*+foo()++++{+return+&A::b;+}+++//+This+is+a+pointer+to+a+member+of+type+int|1042684|从A开始，您可以访问它。|1042685|通过B访问它就是从外部访问它，从而触发访问说明符。|1042686|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

The differece between the two statements becomes more obvious when you try and return the value:

<pre><code>int* foo() { return &amp;(B::b);} // This is a pointer to an int


int A::* foo() { return &amp;B::b; } // This is a pointer to a member of type int
</code></pre>

What you want to do is access it via the A object:

<pre><code>int A::* foo() { return &amp;A::b; } // This is a pointer to a member of type int
</code></pre>

As from the A you are allowed to access it. 
Accessing it via B like that is accessing it from outside and thus triggers the access specifiers.

blocks|key|1870577|text|这只是一个补充。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1870578|§5.3.1/2表示：|1870579|1870580|一元&运算符的结果是指向其操作数的指针。操作数应为左值或限定id。在第一种情况下，如果表达式的类型是“T”，则结果的类型是“指向T的指针”...|blockquote|1870581|1870582|对于合格的id，...如果该成员是类型T的类C的非静态成员，则结果的类型是“指向类型T的类C的成员的指针”。|1870583|1870584|根据§5.1/7，B::b属于限定id的情况，但(B::b)不属于限定id的情况，所以编译器将其解释为左值。|offset|length|style|CODE|1870585|entityMap^0|0|0|0|0|0|0|0|9|4|O|6|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|U|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|V|8|@]|9|@]|A|$]]|$1|E|3|F|5|G|7|W|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|X|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|Y|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|10|8|@$N|11|O|12|P|Q]|$N|13|O|14|P|Q]]|9|@]|A|$]]|$1|R|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|S|$]]

This is just a supplementation. 
§5.3.1/2 says:

<blockquote>
 The result of the unary &amp; operator is
 a pointer to its operand. The operand
 shall be an lvalue or a qualified-id.
 In the first case, if the type of the
 expression is “T,” the type of the
 result is “pointer to T.”
 ... 
 For a
 qualified-id, ... If the member is a
 non-static member of class C of type
 T, the type of the result is “pointer
 to member of class C of type T.”
</blockquote>

According to §5.1/7, <code>B::b</code> comes under the qualified-id case, but <code>(B::b)</code> doesn't.
So, compiler interprets it as an lvalue.

Hello I have this code with a compiler error (error is from Microsoft Visual Studio 2008):
<pre><code>class B {
protected:
 int b;
};

class A : public B {
public:
 void foo() { &amp;B::b; } 
// error C2248: 'B::b' : cannot access protected member declared in class 'B'
};
</code></pre>
while this code is error free:
<pre><code>class B {
protected:
 int b;
};

class A : public B {
public:
 void foo() { &amp;(B::b); }
};
</code></pre>
The two snippets seem to me equivalent based on my knowledge of the precedence of operators, because <code>::</code> has an higher precedence than <code>&amp;</code> (see for example table 2 at page 137 of &quot;<a href="//www.stroustrup.com/JSF-AV-rules.pdf" rel="nofollow noreferrer">JOINT STRIKE FIGHTER AIR VEHICLE C++ CODING STANDARDS FOR THE SYSTEM DEVELOPMENT AND DEMONSTRATION PROGRAM</a>&quot; )
But they are different... I think it is something related to &quot;pointer-to-data-member&quot; but I do not know how does it fit with the operators precedence.
Any explanation?

a question about the precedence of C++ operators "address of" and "scope resolution"

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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你好，我有一个编译器错误的代码(错误来自Microsoft Visual Studio 2008)：class B {protected:    int b;};class A : public B {public:    void foo() { &B::b; } // error C2248: 'B::b' : c...

问关于C++算子“地址”和“作用域归结”优先级的一个问题
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问关于C++算子“地址”和“作用域归结”优先级的一个问题EN